1.  PROGRAM OF “PHYSICS” Lecturer: Dr. DO Xuan Hoi Room 413

2. PHYSICS 2 (FLUID MECHANICS AND THERMAL PHYSICS)
02 credits (30 periods)
Chapter 1 Fluid Mechanics
Chapter 2 Heat, Temperature and the Zeroth Law of Thermodynamics
Chapter 3 Heat, Work and the First Law of Thermodynamics
Chapter 4 The Kinetic Theory of Gases
Chapter 5 Entropy and the Second Law of
Thermodynamics

3. References :
Halliday D., Resnick R. and Walker, J. (2005), Fundamentals of Physics, Extended seventh edition. John Willey and Sons, Inc.
Alonso M. and Finn E.J. (1992). Physics, Addison-Wesley Publishing Company
Hecht, E. (2000). Physics. Calculus, Second Edition. Brooks/Cole.
Faughn/Serway (2006), Serway’s College Physics, Brooks/Cole.
Roger Muncaster (1994), A-Level Physics, Stanley Thornes.

4. .

5. Entropy and the Second Law
Chapter 5
Entropy and the Second Law
of Thermodynamics
1 Heat Engines and
the Second Law of Thermodynamics
2 Reversible and Irreversible Processes
3 The Carnot Engine
4 Gasoline and Diesel Engines
5 Heat Pumps and Refrigerators
6 Entropy

6. 1 Heat Engines and
the Second Law of Thermodynamics
1.1 Notion
A heat engine is a device that converts internal energy (heat) to other useful forms,
such as electrical or mechanical energy
A heat engine carries some working substance through a cyclical process
In internal-combustion engines, such as those used in automobiles, the working substance is a mixture of air and fuel; in a steam turbine it is water.

7. 1.2 Working principle of a heat engine
Energy is transferred from a source at a high temperature (Qh)
Work is done by the engine (Weng)
Energy is expelled to a source at a lower temperature (Qc)
Schematic representation
of a heat engine

8. Since it is a cyclical process, U = 0
Its initial and final internal energies are the same
Therefore, from the first law of thermodynamics :
U = Qnet – Weng = 0
 Qnet = Weng ; Qh – Qc = W
The work done by the engine equals the net energy absorbed by the engine
The work is equal to the area enclosed by the curve of the PV diagram

9. 1.3 Thermal Efficiency of a Heat Engine
Thermal efficiency is defined as the ratio of the work done by the engine to the energy absorbed at the higher temperature
e = 1 (100% efficiency) only if Qc = 0
No energy expelled to cold reservoir

10. 1.4 Second Law of Thermodynamics
“It is impossible to construct a heat engine that, operating in a cycle, produces no other effect than the absorption of energy from a reservoir and the performance of an equal amount of work”
Means that Qc cannot equal 0
Some Qc must be expelled to the environment
Means that e cannot equal 100%
The impossible engine

11. Find the efficiency of a heat engine that absorbs J of energy from a hot reservoir and exhausts
1500 J to a cold reservoir.
PROBLEM 1
SOLUTION

12. A diesel engine performs 2200 J of
mechanical work and discards 4300 J of heat each cycle.
(a) How much heat must be supplied to the engine in each
cycle?
(b) What is the thermal efficiency of the engine?
PROBLEM 2
SOLUTION

13. An aircraft engine takes in 9000 J of heat
and discards 6400 J each cycle.
(a) What is the mechanical work output of the engine during one cycle?
(b) What is the thermal efficiency of the engine?
PROBLEM 3
SOLUTION

14. A gasoline engine has a power output of
180 kW (about 241 hp). Its thermal efficiency is 28.0%.
(a) How much heat must be supplied to the engine per second?
(b) How much heat is discarded by the engine per second?
PROBLEM 4
SOLUTION

15. A gasoline engine in a large truck takes in 10,000 J of heat and delivers 2000 J of mechanical work per cycle. The heat is
obtained by burning gasoline with heat of combustion
LC = 5.0  104 J/g.
(a) What is the thermal efficiency of this engine?
(b) How much heat is discarded in each cycle?
(c) How much gasoline is burned in each cycle?
(d) If the engine goes through 25 cycles per second, what is its
power output in watts? In horsepower?
PROBLEM 5
SOLUTION
(a)
(b)

16. A gasoline engine in a large truck takes in 10,000 J of heat and delivers 2000 J of mechanical work per cycle. The heat is
obtained by burning gasoline with heat of combustion
LC = 5.0  104 J/g.
(a) What is the thermal efficiency of this engine?
(b) How much heat is discarded in each cycle?
(c) How much gasoline is burned in each cycle?
(d) If the engine goes through 25 cycles per second, what is its
power output in watts? In horsepower?
PROBLEM 5
SOLUTION
(c)
(d)

17. 2 Reversible and Irreversible Processes
Thermodynamic processes that occur in nature are all irreversible processes.
These are processes that proceed spontaneously
in one direction but not the other
EXAMPLE :
 The flow of heat from a hot body to a cooler body
 Sliding a book across a table converts mechanical energy into heat by friction
(a book initially at rest on the table would spontaneously start moving and the table and book would cool down ?).
 A block of ice melts irreversibly when we place it in a hot (70°C) metal box.

18. 2 Reversible and Irreversible Processes
A reversible process is one in which every state along some path is an equilibrium state
And one for which the system can be returned to its initial state along the same path
An irreversible process does not meet these requirements
Most natural processes are irreversible
Reversible process are an idealization, but some real processes are good approximations
Reversible processes are thus equilibrium processes, with the system always in thermodynamic equilibrium.

19. 3 THE CARNOT ENGINE A theoretical engine developed by Sadi Carnot
A heat engine operating in an ideal, reversible cycle (now called a Carnot Cycle) between two reservoirs is the most efficient engine possible
Carnot’s Theorem : No real engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs

20. Carnot Cycle, A to B A to B is an isothermal expansion
The gas is placed in contact with the high temperature reservoir
The gas absorbs heat Qh
The gas does work WAB in raising the piston

21. Carnot Cycle, B to C B to C is an adiabatic expansion
The base of the cylinder is replaced by a thermally nonconducting wall
No heat enters or leaves the system
The temperature falls from Th to Tc
The gas does work WBC

22. Carnot Cycle, C to D
The gas is placed in contact with the cold temperature reservoir
C to D is an isothermal compression
The gas expels energy QC
Work WCD is done on the gas

23. Carnot Cycle, D to A D to A is an adiabatic compression
The gas is again placed against a thermally nonconducting wall
So no heat is exchanged with the surroundings
The temperature of the gas increases from TC to Th
The work done on the gas is WCD

24.  The Camot cycle consists of two reversible isothermal and two reversible adiabatic processes.

25. Carnot Cycle, PV Diagram
The work done by the engine is shown by the area enclosed by the curve
The net work is equal to Qh - Qc
The thermal efficiencyof the engine :
We can demonstrate :
 The thermal efficiency of a Carnot engine :
(The efficiency of a Carnot engine depends only on the
temperatures of the two heat reservoirs)

26. Notes About Carnot Efficiency
Efficiency is 0 if Th = Tc
Efficiency is 100% only if Tc = 0 K
Such reservoirs are not available
The efficiency increases at Tc is lowered and as Th is raised
In most practical cases, Tc is near room temperature, 300 K
So generally Th is raised to increase efficiency
All real engines are less efficient than the Carnot engine
Real engines are irreversible because of friction
Real engines are irreversible because they complete cycles in short amounts of time

27. A steam engine has a boiler that operates at
500 K. The energy from the burning fuel changes water to
steam, and this steam then drives a piston. The cold
reservoir’s temperature is that of the outside air,
approximately 300 K. What is the maximum thermal efficiency
of this steam engine?
PROBLEM 6
SOLUTION

28. The highest theoretical efficiency of a certain
engine is 30%.
If this engine uses the atmosphere, which has a temperature
of 300 K, as its cold reservoir, what is the temperature of its
hot reservoir?
PROBLEM 7
SOLUTION

29. A Carnot engine takes 2000 J of heat from a
reservoir at 500 K, does some work, and discards some heat
to a reservoir at 350 K. How much work does it do, how much
heat is discarded, and what is the efficiency ?
PROBLEM 8
SOLUTION
(Qc < 0 : heat flows out of the engine and into the cold reservoir.)

30. Suppose 0.200 mol of an ideal diatomic gas (
= 1.40) undergoes a Carnot cycle with temperatures 227°C
and 27°C. The initial pressure is Pa = 10.0  105 Pa, and
during the isothermal expansion at the higher temperature
the volume doubles. (a) Find the pressure and volume at each
of points a, b, c, and d in the P V-diagram of the figure.
PROBLEM 9
SOLUTION

31. Suppose 0.200 mol of an ideal diatomic gas (
= 1.40) undergoes a Carnot cycle with temperatures 227°C
and 27°C. The initial pressure is Pa = 10.0  105 Pa, and
during the isothermal expansion at the higher temperature
the volume doubles. (a) Find the pressure and volume at each
of points a, b, c, and d in the P V-diagram of the figure.
PROBLEM 9
SOLUTION

32. Suppose 0.200 mol of an ideal diatomic gas (
= 1.40) undergoes a Carnot cycle with temperatures 227°C
and 27°C. The initial pressure is Pa = 10.0  105 Pa, and
during the isothermal expansion at the higher temperature
the volume doubles. (b) Find Q, W, and U for each step
and for the entire cycle.
PROBLEM 9
SOLUTION

33. Suppose 0.200 mol of an ideal diatomic gas (
= 1.40) undergoes a Carnot cycle with temperatures 227°C
and 27°C. The initial pressure is Pa = 10.0  105 Pa, and
during the isothermal expansion at the higher temperature
the volume doubles. (b) Find Q, W, and U for each step
and for the entire cycle.
PROBLEM 9
SOLUTION

34. Suppose 0.200 mol of an ideal diatomic gas (
= 1.40) undergoes a Carnot cycle with temperatures 227°C
and 27°C. The initial pressure is Pa = 10.0  105 Pa, and
during the isothermal expansion at the higher temperature
the volume doubles. (b) Find Q, W, and U for each step
and for the entire cycle.
PROBLEM 9
SOLUTION

35. Suppose 0.200 mol of an ideal diatomic gas (
= 1.40) undergoes a Carnot cycle with temperatures 227°C
and 27°C. The initial pressure is Pa = 10.0  105 Pa, and
during the isothermal expansion at the higher temperature
the volume doubles. (c) Determine the efficiency directly from
the results of part (b).
PROBLEM 9
SOLUTION

36. 4 Internal-Combustion Engines
 Gasoline and Diesel Engines
(used in automobiles and many other types of machinery) are familiar examples of a heat engine.
a. Gasoline Engines : In a gasoline engine, six processes occur in each cycle

37. Cycle of a four-stroke internal-combustion engine.
Intake stroke: Piston moves down, causing a partial vacuum in cylinder; gasoline-air mixture enters through intake valve.

38. Cycle of a four-stroke internal-combustion engine.
Compression stroke: Intake valve closes; mixture is compressed as piston moves up.
Ignition: Spark plug ignites mixture.
Power stroke: Hot burned mixture expands, pushing piston down.

39. Cycle of a four-stroke internal-combustion engine.
Exhaust stroke: Exhaust valve opens; piston moves up, expelling exhaust and leaving cylinder ready for next intake stroke.

40. Cycle of a four-stroke internal-combustion engine.

41. The Otto cycle
The Otto cycle is an idealized model of the thermodynamic
processes in a gasoline engine
If the air–fuel mixture is assumed to be an ideal gas, then the efficiency of the Otto cycle is
where  is the ratio of the molar specific heats :
and V1/V2 is the compression ratio
NOTES :  For a typical compression ratio of 8 and with
 = 1.4, we predict a theoretical efficiency of 56%
 Efficiencies of real gasoline engines are typically around 35%

42. b. The Diesel Cycle
 Diesel engines operate on a cycle similar to the Otto cycle but do not employ a spark plug. There is no fuel in the cylinder at the beginning of the compression stroke.
 The compression ratio for a diesel engine is much greater than that for a gasoline engine
 Diesel engines are more efficient than gasoline engines because of their greater compression ratios and resulting higher combustion temperatures.

43. What compression ratio r must an Otto cycle
have to achieve an ideal efficiency of 65.0% if  = 1.40 ?
PROBLEM 10
SOLUTION
NOTES : e increases when r increases.

44. (a) Calculate the theoretical efficiency for an
Otto cycle engine with  = 1.40 and r = 9.50.
(b) If this engine takes in 10,000 J of heat from burning its
fuel, how much heat does it discard to the outside air ?
PROBLEM 11
SOLUTION
(a)
(b)

45. 5 Heat Pumps and Refrigerators
 In a refrigerator or heat pump, the
engine absorbs energy Qc from a cold reservoir and expels energy Qh to a hot reservoir
some work must be done on the engine
Refrigerator
Impossible refrigerator

46.  The principles of the common refrigeration cycle
The fluid "circuit" contains a refrigerant fluid (the working substance)
The compressor takes in fluid, compresses it adiabatically, and delivers it to the condenser coil at high pressure.
the refrigerant gives off heat QH and partially condenses to liquid
The fluid then expands adiabatically into the evaporator
As the fluid expands, it cools considerably, enough that the fluid in the evaporator coil is colder than its surroundings.

47.  Effectiveness of a refrigerator: coefficient of performance (COP)
QC > W  COP >1
It is desirable for the COP to be as high as possible
Refrigerator
To compare with the thermal efficiency of a Heat Engine:

48. A refrigerator has a coefficient of
performance of In each cycle it absorbs 3.40104 J of
heat from the cold reservoir.
(a) How much mechanical energy is required each cycle to
operate the refrigerator?
(b) During each cycle, how much heat is discarded to the
high-temperature reservoir?
PROBLEM 12
SOLUTION

49. 6 Entropy
 Pressure and volume are state properties, properties that depend only on the state of the gas and not on how it
reached that state. Other state properties are temperature and energy.
 Another state property: its entropy.
 Entropy S of the system is defined by :
J/K
(The change in entropy of a system during an arbitrary
process between an initial state i and a final state f )

50.  To find the entropy change for an irreversible process occurring in a closed system, replace that process with any reversible process that connects the same initial and final states.
The isothermal expansion:
To keep the temperature T of the gas constant during the isothermal expansion: heat Q must have been energy transferred from the reservoir to the gas
 Q is positive and the entropy of the gas increases during the isothermal process and during the free expansion

51.  Entropy as state function
To make the process reversible, it is done slowly in a series of small steps, with the gas in an equilibrium state at the end of each step. For each small step, the energy transferred as heat to or from the gas is dQ, the work done by the gas is dW, and the change in internal energy is dEint.
The first law of thermodynamics:

52.  The second law of thermodynamics :
“The total entropy of an isolated system that undergoes a change can never decrease”
 The entropy of the Universe increases in all real processes
 Entropy provides a quantitative measure of disorder.

53. Example
Consider the changes in entropy that occur in a Carnot heat engine operating between the temperatures Tc and Th .
But for Carnot cycle :

54.  Consider a system taken through an arbitrary (non-Carnot) reversible cycle
Entropy S is a state function  S depends only on
the properties of a given equilibrium state  S = 0
 When a system is in a particular macroscopic state, the particles that make up the system may be in any of w possible microscopic states.
From a microscopic viewpoint, entropy is defined as :
k : Boltzmann’s constant;
W : the number of microstates available to the system

55. One kilogram of ice at O°C is melted and
converted to water at O°C. Compute its change in entropy,
assuming that the melting is done reversibly. The heat of
fusion of water is Lf = 3.34  105 J/kg.
PROBLEM 13
SOLUTION
The heat needed to melt the ice :

56. Suppose 1.0 mol of nitrogen gas is confined
to the left side of the container. You open the stopcock, and
the volume of the gas doubles.
What is the entropy change of the gas for this irreversible
process? Treat the gas as ideal.
PROBLEM 14
SOLUTION
The entropy change for the irreversible process by calculating it for a reversible process that provides the same change in volume. The temperature of the gas does not change  isothermal expansion

57. One kilogram of water at 0°C is heated to
100°C. Compute its change in entropy.
Specific heat of water is 4190 J/kg.K
PROBLEM 15
SOLUTION

58. Suppose 1.00 kg of water at 100°C is placed
in thermal contact with 1.00 kg of water at 0°C. What is the
total change in entropy? Assume that the specific heat of
water is constant at 4190 J/kg.K over this temperature range.
PROBLEM 16
SOLUTION

59. 4 Microscopic Interpretation of Entropy
4.1 Macroscopic and microscopic states
 Toss 4 identical coins on the floor:
There can be many microscopic states that correspond to the same macroscopic description.
N coins that are all heads constitute a completely ordered macroscopic state
(macrostate)
 The macroscopic description "half heads, half tails" : the system is disordered
 The state "half heads, half tails"
has a much greater number of possible microscopic
states (microstates), much greater disorder :
It is the most probable

60. 4.2 Microscopic expression for entropy
 If entropy is a quantitative measure of disorder
For any system, the most probable macrostate is the one with the greatest number of corresponding microstates, which is also the macrostate with the greatest disorder and the greatest entropy.
 Let w represent the number of possible microscopic states
(or multiplicity) for a given macrostate. 
the entropy of a macroscopic state :

61.  process that takes it from macrostate 1, for which
 Consider a system that undergoes a thermodynamic
process that takes it from macrostate 1, for which
there are w1 possible microstates (multiplicity w1), to
macroscopic state 2, with w2 associated microstates
(multiplicity w2) .
The change in entropy in this process is 
The difference in entropy between the two macrostates depends on the ratio of the numbers of possible
microstates.

62. between the two halves of a box.
 For a system of N molecules that may be distributed
between the two halves of a box.
n1: the number of molecules in one half of the box and
n2 : the number in the other half.
The multiplicity:
When N is very large (the usual case):
(Stirling's approximation)

63. Suppose that there are 1-00 indistinguishable
molecules in the box.
How many microstates are associated with the configuration
(n1 = 50; n2 = 50),
and with the configuration (n1 = 100; n2 = 0)?
PROBLEM 17
SOLUTION
 Configuration (n1 = 50; n2 = 50)
The multiplicity:
 Configuration (n1 = 100; n2 = 0)
The multiplicity:

64.  Initial configuration (n1 = 100; n2 = 0)
A thermally insulated box is divided by a partition
into two compartments, each having volume V. Initially, one com-
partment contains n moles of an ideal gas at temperature T, and
the other compartment is evacuated. We then break the partition,
and the gas expands to fill both compartments. What is the entropy
change in this free-expansion process?
PROBLEM 18
SOLUTION
NB: Use of the equation
 Initial configuration (n1 = 100; n2 = 0)
Initial multiplicity:
Initial entropy:

65.  Final configuration (n1 = N/2; n2 = N/2)
A thermally insulated box is divided by a partition
into two compartments, each having volume V. Initially, one com-
partment contains n moles of an ideal gas at temperature T, and
the other compartment is evacuated. We then break the partition,
and the gas expands to fill both compartments. What is the entropy
change in this free-expansion process?
PROBLEM 18
SOLUTION
 Final configuration (n1 = N/2; n2 = N/2)
Final multiplicity:
Final entropy:
The change in entropy:

66. A thermally insulated box is divided by a partition
into two compartments, each having volume V. Initially, one com-
partment contains n moles of an ideal gas at temperature T, and
the other compartment is evacuated. We then break the partition,
and the gas expands to fill both compartments. What is the entropy
change in this free-expansion process?
PROBLEM 18
SOLUTION
When the partition is broken, each molecule now
has twice as much volume in which it can move
and hence has twice the number of possible positions.
Let w1 be the number of microscopic states of the system as a whole when the gas occupies volume V
The number w2 of microscopic states when the gas occupies volume 2V is greater by a factor of 2N : w2 = 2Nw1

67. V. The gas expands isothermally and reversibly to a volume
PROBLEM 19
Two moles of an ideal gas occupy a volume
V. The gas expands isothermally and reversibly to a volume
3V. (a) Is the velocity distribution changed by the isothermal
expansion? (b) Calculate the change in entropy of the gas.
SOLUTION
(a) The velocity distribution depends only on T, so in an isothermal process it does not change.
(b) The number of possible positions available to each molecule is
altered by a factor of 3 (becomes larger).
Hence the number of microscopic states the gas occupies at
volume 3V is w2= (3)Nw1, where N is the number of molecules
and w1 is the number of possible microscopic states at the start
of the process.